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For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is

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anandsebastin
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For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink] For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is   Updated on: 24 Oct 2006, 14:36
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For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is the value of a?

1. a^2 - a = 12
2. b^2 - b = 2

Please explain your answers.

(Edit: Added the original question for clarity.)



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Originally posted by anandsebastin on 24 Oct 2006, 13:53.
Last edited by anandsebastin on 24 Oct 2006, 14:36, edited 1 time in total.
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mst
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Re: For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink] For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is   24 Oct 2006, 14:09
I get C.

From 1, a = 4 or -3
From 2, b = 2 or -1.

sqrt ( a^3 - a^2 - b) = 7 is satisfied only by a = 4 and b = -1
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Fig
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Re: For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink] For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is   24 Oct 2006, 14:15
for me (C) :)

sqrt ( a^3 - a^2 - b) = 7
=> a^3 - a^2 - b = 49 (1)

Stat 1 :

a^2 - a = 12
<=> a^2 - a - 12 = 0
<=> (a+3)(a-4) = 0
<=> a = -3 or a = 4

Both values of a work in (1)

INSUFF

Stat 2 :

b^2 - b = 2
<=> b^2 - b - 2 = 0
<=> (b-2)*(b+1) = 0
<=> b=2 or b=-1

Placed in (1), we have several possible values of a

INSUFF

Combining (1) with (2):

case 1 : a = -3
(1) => (-3)^3 - (-3)^2 -b = 49
<=> -36 - b = 49
<=> b = -85 >>> Not matching values of b in statment 2

case 2 : a = 4
(1) => (4)^3 - (4)^2 -b = 49
<=> 64 - 16 - b = 49
<=> b = 48-49 = -1 >>> Bingo :) 1 match with 1 value of b in statment 2
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Re: For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink] For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is   25 Oct 2006, 11:42
anandsebastin wrote:
For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is the value of a?

1. a^2 - a = 12
2. b^2 - b = 2




B
as mst and fig have calculated
from stat 2, b = 2 or b=-1

from stem a^3-a^2-b=49
hence a^3-a^2 = 49+b
a*a*(a-1) = 49+b

if b = 2
a*a*(a-1) = 51
but 51 = 17*3 and neither of them can be used as a in equation above (and we have to find integer a)

if b=-1
a*a*(a-1) = 48

factoring 48, it is possible to get
48 = 4*4*3
hence a=4

from stat 1, a=4 or a=-3
both can be used to satisfy the stem equation and arrive at different solutions for b as b = a*a*(a-1) -49 where b= -1 or b=-85
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Re: For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink] For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is   25 Oct 2006, 13:38
I'm getting B as well
the only option for a and b to be integers is if a=4 and b=-1
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Re: For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink] For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is   25 Oct 2006, 13:55
Yesss :)..... It's (B) :)

This is my favorite silly mistake :) : to forget that we are speaking about integers. As a is in integer, there is only 1 possible solution with the statment 2 :)

Thanks guys :D :)
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Re: For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink] For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is   25 Oct 2006, 18:38
Muddypaws nailed it.

OA is B. The explanation was scrambled in Princeton Review.



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Re: For integers, a and b, if sqrt ( a^3 - a^2 - b) = 7, what is [#permalink]
25 Oct 2006, 18:38
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